Formatted question description: https://leetcode.ca/all/646.html

646. Maximum Length of Pair Chain (Medium)

You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.

Now, we define a pair (c, d) can follow another pair (a, b) if and only if b < c. Chain of pairs can be formed in this fashion.

Given a set of pairs, find the length longest chain which can be formed. You needn't use up all the given pairs. You can select pairs in any order.

Example 1:

Input: [[1,2], [2,3], [3,4]]
Output: 2
Explanation: The longest chain is [1,2] -> [3,4]

Note:

The number of given pairs will be in the range [1, 1000].

Related Topics:
Dynamic Programming

Similar Questions:

Solution 1. DP

First sort the array in ascending order.

Let dp[i] be the length of maximum chain formed using a subsequence of A[0..i] where A[i] must be used.

dp[i] = max(1, max(1 + dp[j] | pair j can go after pair i ))

// OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int findLongestChain(vector<vector<int>>& A) {
        sort(begin(A), end(A));
        int N = A.size();
        vector<int> dp(N, 1);
        for (int i = 1; i < N; ++i) {
            for (int j = 0; j < i; ++j) {
                if (A[j][1] >= A[i][0]) continue;
                dp[i] = max(dp[i], 1 + dp[j]);
            }
        }
        return *max_element(begin(dp), end(dp));
    }
};

Solution 3. Interval Scheduling Maximization (ISM)

// OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int findLongestChain(vector<vector<int>>& A) {
        sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
        int e = INT_MIN, ans = 0;
        for (auto &v : A) {
            if (e >= v[0]) continue;
            e = v[1];
            ++ans;
        }
        return ans;
    }
};

Java

class Solution {
    public int findLongestChain(int[][] pairs) {
        Arrays.sort(pairs, new Comparator<int[]>() {
            public int compare(int[] pair1, int[] pair2) {
                if (pair1[0] != pair2[0])
                    return pair1[0] - pair2[0];
                else
                    return pair1[1] - pair2[1];
            }
        });
        int length = pairs.length;
        int[] dp = new int[length];
        Arrays.fill(dp, 1);
        for (int i = 1; i < length; i++) {
            int[] curPair = pairs[i];
            for (int j = 0; j < i; j++) {
                int[] prevPair = pairs[j];
                if (prevPair[1] < curPair[0])
                    dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
        int max = 1;
        for (int num : dp)
            max = Math.max(max, num);
        return max;
    }
}

// OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int findLongestChain(vector<vector<int>>& A) {
        sort(begin(A), end(A));
        int N = A.size();
        vector<int> dp(N, 1);
        for (int i = 1; i < N; ++i) {
            for (int j = 0; j < i; ++j) {
                if (A[j][1] >= A[i][0]) continue;
                dp[i] = max(dp[i], 1 + dp[j]);
            }
        }
        return *max_element(begin(dp), end(dp));
    }
};

class Solution(object):
  def findLongestChain(self, pairs):
    """
    :type pairs: List[List[int]]
    :rtype: int
    """
    tails = []
    for start, end in sorted(pairs):
      idx = bisect.bisect_left(tails, start)
      if idx == len(tails):
        tails.append(end)
      else:
        tails[idx] = min(tails[idx], end)
    return len(tails)

646 - Maximum Length of Pair Chain

646. Maximum Length of Pair Chain (Medium)

Solution 1. DP

Solution 3. Interval Scheduling Maximization (ISM)

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